S.chand publication solution of class 8 chapter 2 exercise 2 Exponents

EXERCISE 2

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Q1 | Ex-2 | Class 8 | Schand composite mathematics solution | myhelper

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Question 1

Evaluate:

(i) 2^-3

Sol :

Using property $a^{-n}=\dfrac{1}{a^n}$

⇒$\dfrac{1}{2^3}=\dfrac{1}{2}\times \dfrac{1}{2} \times \dfrac{1}{2}$

⇒$\dfrac{1}{2\times 2 \times 2}=\dfrac{1}{8}$

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(ii) 6⁰

Sol :

Using property: a⁰=1

⇒6⁰=1

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(iii) (-3)^-2

Sol :

Using property $a^{-n}=\dfrac{1}{a^n}$

⇒$\dfrac{1}{(-3)^2}=\dfrac{1}{-3}\times \dfrac{1}{-3}$

⇒$\dfrac{1}{-3\times -3 }=\dfrac{1}{9}$

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(iv) 8⁵ . 8^-5

Sol :

Using property a^m×aⁿ=a^(m+n)

⇒8⁵ . 8^-5 or

⇒8⁵ × 8^-5 

⇒8^(5-5)

⇒8⁰

⇒1

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(v) $\dfrac{1}{2^{-2}}$

Sol :

Using property $\dfrac{1}{a}=a^{-1}$

⇒2^(-2)×-1

⇒2²

⇒4

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(vi) 3(8-1)⁰

Sol :

Using property: a⁰=1

⇒3×1

⇒3

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(vii) $\dfrac{1}{2^{-1}}+\dfrac{1}{3^{-1}}$

Sol :

Using property : $\dfrac{1}{a^{-1}}=a$

⇒2+3

⇒5

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(viii) 7x⁰

Sol :

Using property: a⁰=1

⇒7×1

⇒7

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(ix) a² ÷ a^-2

Sol :

Using property : $a^{-n}=\dfrac{1}{a^n}$

⇒$a^2 \div \dfrac{1}{a^2}$

⇒$a^2 \times \dfrac{a^2}{1}$

⇒a²×a²

Using property : aⁿ×a^m=a^(n+m)

⇒a⁴

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(x) $\dfrac{y^{-6}}{y^4}$

Sol :

Using property : $a^{-n}=\dfrac{1}{a}$

⇒$\dfrac{1}{(y^4)\times (y^6)}$

Using property: a^m×aⁿ=a^(m+n)

⇒$\dfrac{1}{y^{(6+4)}}=\dfrac{1}{y^{(10)}}$ or

⇒y^(-10)

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(xi) (x ^-3)⁰

Sol :

Using property: (a^m)ⁿ=a(m×n)

⇒x⁰

⇒1

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(xii) 6⁰ + 6 ^-1 + 3 ^-1

Sol :

Using property: a⁰=1

⇒1+6 ^-1 + 3 ^-1

Using property: $a^{-1}=\dfrac{1}{a}$

⇒$1+\dfrac{1}{6}+\dfrac{1}{3}$

L.C.M of 6 and 3 is =2×3=6

$\begin{array}{l|l}

2&6,3 \\ \hline

3&3,3 \\ \hline

&1,1 \end{array}$

⇒$\dfrac{1\times 6+1\times 1+1\times 2}{6}=\dfrac{6+1+2}{6}$

or $\dfrac{9}{6}=\dfrac{3}{2}$

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(xiii) $\dfrac{1}{a^0 + b^0}$

Sol :

Using property: a⁰=1

⇒$\dfrac{1}{1+ 1}=\dfrac{1}{2}$

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(xiv) (3x²)³

Sol :

Using property: (aⁿ)^m=a^(m×n)

⇒ 3⁽³⁾× x^(2×3)

⇒ 27× x⁶

⇒ 27x⁶

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(xv) $\left[\left(\dfrac{5}{7}\right)^{-6}\right]^{-0}$

Sol :

Using property: (a^m)ⁿ=a^(m×n)

⇒$\left(\dfrac{5}{7}\right)^{-6\times -0}$

⇒$\left(\dfrac{5}{7}\right)^0$

Using property : a⁰=1

⇒1

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Question 2

Simplify:

(i) x⁵ . x³

Sol :

Using property : a^m×aⁿ=a^(m+n)

⇒x⁽⁵⁺³⁾

⇒x⁽⁸⁾

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(ii) $\dfrac{x^8}{x^3}$

Sol :

Using property : $\dfrac{1}{a^n}=a^{-n}$

⇒x⁸ × x^(-3)

Using property : a^m×aⁿ=a^(m+n)

⇒x^(8-3)

⇒x⁵

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(iii) (x³)²

Sol :

Using property : a^m×aⁿ=a^(m+n)

⇒x^(3×2)

⇒x⁶

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(iv) (3x²)³

Sol :

Using property : a^m×aⁿ=a^(m+n)

⇒3³×(x²)³

⇒27×x⁶

⇒27x⁶

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(v) (5x^-3yz²)^-2

Sol :

Using property : $a^{-n}=\dfrac{1}{a^n}$

⇒$\dfrac{1}{(5x^{-3}yz^2)^2}$

⇒$\dfrac{1}{5^2 \times x^{-3\times 2}\times y^2 z^{2\times 2}}$

⇒$\dfrac{1}{25 \times x^{-6}\times y^2 \times z^{4}}$

⇒$\dfrac{x^6}{25y^2 z^{4}}$

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Question 3

Write the option that indicates the correct meaning of the given expression

(i) 4n³ means

(a) 4 times n + n + n

(b) 4n . 4n . 4n

(c) 4 . n . n . n

Sol :

⇒4n³

⇒4.n.n.n or (4×n×n×n)

Option (c) is correct

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(ii) 6x³y² means

(a) 6 . xy . xy . xy . xy . xy

(b) (6 . x. x. x . y)(7. x. x. x . y)

(c) 6 . x . x . x . y .y

Sol :

Option (c) is correct

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Question 4

Simplify:

(i) $\dfrac{(a^0b^{-2})^5}{2a^{-1}}$

Sol :

⇒$\dfrac{(a^{0\times 5}b^{-2\times 5})}{2a^{-1}}$

⇒$\dfrac{(a^{0}b^{-10})}{2a^{-1}}$

Using property: $\dfrac{1}{a^{-1}}=a$

⇒$\dfrac{a^0 \times a^1}{2b^{10}}$

Using property: a⁰=1

⇒$\dfrac{a^1}{2b^{10}}$

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(ii) $\dfrac{4a^{-3}b^{0}}{2a^{2}b^{-1}}$

Sol :

⇒$\dfrac{\not{4}a^{-3}b^{0}}{\not{2}a^{2}b^{-1}}$

Using property: $\dfrac{1}{a}=a^{-1}$

⇒2a^-3×b⁰× a^-2 × b⁺¹

Using property: aⁿ×a^m=a^(m+n)

⇒2a^-3-2×b⁰⁺¹

⇒2a^-5×b¹

⇒$\dfrac{2b}{a^5}$

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(iii) $\dfrac{2x^{2}y^{-2}}{2^{-1}x^{2}y^{2}}$

Sol :

Using property : $\dfrac{1}{a^{-1}}=a^1$

⇒(2x²y^-2)(2¹x^-2y^-2)

⇒(2×2)x^2-2y^-2-2

⇒4x⁰y^-4

Using property : a⁰=1

⇒4y^-4

⇒$\dfrac{4}{y^4}$

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(iv) $\dfrac{(t^{-4})^3}{(t^{3})^{-4}}$

Sol :

Using property: (a^m)ⁿ=a^(m×n)

⇒$\dfrac{t^{-12}}{t^{-12}}$

⇒1

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(v) 4k²(4^-1k+4k^-2)

Sol :

⇒4k²(4^-1k)+4k²(4k^-2)

using property: $\dfrac{1}{a}=a^{-1}$

⇒$\dfrac{4k^{2}\times k}{4}+\dfrac{4k^{2}\times 4}{k^{2}}$

⇒$\dfrac{k^{2+1}}{1}+\dfrac{16k^{2}}{k^{2}}$

⇒k³+16

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(vi) $\left(\dfrac{3x^{-2}}{2y^{-1}}\right)^{-2}$

Sol :

⇒$\dfrac{(3x^{-2})^{-2}}{(2y^{-1})^{-2}}$

Using property : (a^m)ⁿ=a^(m×n)

⇒$\dfrac{3^{-2}x^{4})}{2^{-2}y^{2}}$

⇒$\dfrac{9^{-1}x^{4})}{4^{-1}y^{2}}$

⇒$\dfrac{4x^{4})}{9y^{2}}$

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(vii) $\left[\dfrac{3^{-1}}{(-2)^{-2}}\right]^{-2}$

Sol :

⇒$\dfrac{(3^{-1})^{-2}}{\{(-2)^{-2}\}^{-2}}$

Using property : (a^m)ⁿ=a^(m×n)

⇒$\dfrac{3^{2}}{(-2)^{4}}=\dfrac{9}{16}$

 

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Question 5

Express each of the following as a power of 2 : 8 , 8^x , 16^x+3

Sol :

⇒8=2×2×2=2³

⇒8^x=(2×2×2)^x=2^3x

⇒16^x+3=(2×2×2×2)^x+3

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Question 6

If a = 2^m and b = 2^{m+1} , show that $\dfrac{8a^3}{b^2}=2^{m+1}$

Sol :

⇒$\dfrac{8a^3}{b^2}$

⇒$\dfrac{8(2^m)^3}{(2^{m+1})^2}$

Using property : (a^m)ⁿ=a^(m×n)

⇒$\dfrac{2^3\times 2^{3m}}{2^{2m+2}}$

Using property : a^(m×n)=(a^m)ⁿ

⇒$\dfrac{2^3\times 2^{2m+m}}{2^{2m}+2^{2}}$

⇒$\dfrac{2^3\times 2^{2m}\times 2^{m}}{2^{2m} \times 2^{2}}$

⇒$\dfrac{2^3\times \not{(2^{2m} )}\times 2^{m}}{\not{(2^{2m}) }\times 2^{2}}$

⇒2×2^m

Using property : a^m×aⁿ=a^(m+n)

⇒2^m+1

Hence proved

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Question 7

If x = 2^k and y = 2^k+3 , what is the value of $\dfrac{x}{y}$ ?

Sol :

⇒$\dfrac{x}{y}$

⇒$\dfrac{2^k}{2^{k+3}}$

Using property: a^(m+n)=a^m×aⁿ

⇒$\dfrac{\not{(2^k)}}{\not{(2^{k})}+2^{3}}$

⇒$\dfrac{1}{8}$

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Question 8

Write each expression such that there is no variable in the denominator .

(i) $\dfrac{3r^{-2}s^{3}}{12r^{-3}s^{7}}$

Sol :

⇒(3r^-2s³)(r³s^-7)/12

Using property: a^m×aⁿ=a^(m+n)

⇒$\dfrac{3}{12}(r^{-2+3}s^{3-7})$

⇒$\dfrac{1}{4}(r^{1}s^{-4})$

⇒$\dfrac{r^{1}s^{-4}}{4}$

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(ii) $\dfrac{12x^3y^{-2}z^{4}}{6x^{7}y^{-5}z^{-3}}$

Sol :

⇒$\dfrac{12}{6}(x^3 y^{-2} z^4)(x^{-7} y^{5} z^{3})$

Using property: a^m×aⁿ=a^(m+n)

⇒2(x^3-7×y^-2+5×z⁴⁺³)

⇒2(x^-4×y³×z⁷)

⇒2x^-4y³z⁷

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(iii) $\dfrac{(5m)^{0}n^{-2}}{4mn^{-3}}$

Sol :

⇒$\dfrac{5^0}{4}(m^0 \times n^{-2})(m^{-1}n^{3})$

Using property: a⁰=1

⇒$\dfrac{1}{4}(1 \times n^{-2})(m^{-1}n^{3})$

Using property: a^m×aⁿ=a^(m+n)

⇒$\dfrac{1}{4}( n^{-2+3})(m^{-1})$

⇒$\dfrac{nm^{-1}}{4}$

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(iv) $\dfrac{14a^{-4}k}{7^{0}a^{3}k^{8}}$

Sol :

⇒$\dfrac{14}{7^0}(a^{-4}k)(a^{-3}k^{-8})$

Using property: a⁰=1

⇒14 (a^-4k)(a^-3k^-8)

⇒14 (a^-4-3k^1-8)

⇒14 (a^-7k^-7)

 

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Question 9

Simplify:

(i) $\dfrac{2^{-n}. 8^{2n-1} . 16^{2n}}{4^{3n}}$

Sol :

⇒$\dfrac{2^{-n}\times 2^{3(2n-1)} \times 2^{4(2n)}}{2^{2(3n)}}$

⇒$\dfrac{2^{-n}\times 2^{6n-3} \times 2^{8n}}{2^{6n}}$

⇒$\dfrac{2^{-n}\times \not{(2^{6n})} \times 2^{-3} \times 2^{8n}}{\not{(2^{6n})}}$

⇒2^-n× 2^-3× 2⁸ⁿ

Using property : a^m×aⁿ=a^(m+n)

⇒2^-n-3+8n

⇒2^+7n-3

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(ii) $\dfrac{2^{n+4}-2.2^{n}}{2.2^{n+3}}+2^{-3}$

Sol : 

Using property : a^m×aⁿ=a^(m+n)

⇒$\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}+2^{-3}$

⇒$\dfrac{\not{(2^{n+4})}}{\not{(2^{n+4})}}-\dfrac{2^{n+1}}{2^{n+4}}+2^{-3}$

⇒$1-\dfrac{2^{n+1}}{2^{(n+1)+(3)}}+2^{-3}$

⇒$1-\dfrac{\not{(2^{n+1})}}{\not{(2^{(n+1)})}\times 2^{(3)}}+2^{-3}$

⇒$1-\dfrac{1}{2^3}+\dfrac{1}{2^3}$

⇒1

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(iii)  $\dfrac{(0.6)^0-(0.1)^{-1}}{\left(\dfrac{3}{2^{3}}\right)^{-1}.\left(\dfrac{3}{2}\right)^{3}+\left(-\dfrac{1}{3}\right)^{-1}}$

Sol :

Using property: a⁰=1

⇒$\dfrac{1-\left(\dfrac{1}{10}\right)^{-1}}{\left(\dfrac{3}{2^{3}}\right)^{-1} \times \left(\dfrac{3}{2}\right)^{3} + \left(-\dfrac{1}{3}\right)^{-1}}$

Using property: $a^{-1}=\dfrac{1}{a}$

⇒$\dfrac{1-\dfrac{10}{1}}{\left(\dfrac{2^{3}}{3}\right) \times \left(\dfrac{3^3}{2^3}\right) - \left(\dfrac{3}{1}\right)}$

⇒$\dfrac{1-10}{\left(\dfrac{\not{(8)}}{\not{(3)}}\right) \times \left(\dfrac{\not{(27)}}{\not{(8)}}\right) -3}$

⇒$\dfrac{-9}{9-3}=-\dfrac{9}{6}=-\dfrac{3}{2}$

 

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Question 10

If $\dfrac{9^n \times 3^5 \times (27)^3}{3\times (81)^4}=27$ , then find the value of n

Sol :

⇒$\dfrac{9^n \times 3^5 \times (27)^3}{3\times (81)^4}$

⇒$\dfrac{3^{2n} \times 3^5 \times 3^{9}}{3^{1}\times 3^{16}}$

Using property: a^m×aⁿ=a^(m+n)

⇒$\dfrac{3^{2n+5+9}}{3^{16+1}}$

⇒$\dfrac{3^{2n+14}}{3^{17}}$

⇒3²ⁿ×3¹⁴×3^-17

⇒3²ⁿ×3^14-17

⇒3²ⁿ×3^-3

⇒$\dfrac{3^{2n}}{3^3}=27$  according to question

⇒3²ⁿ=3³×3³

⇒3²ⁿ=3⁶

⇒3²ⁿ=3^2×3

⇒n=3

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Question 11

Show that :

(i) $\left(\dfrac{x^a}{x^{b}}\right)^{(a+b)} \times \left(\dfrac{x^b}{x^c}\right)^{(b+c)} \times \left(\dfrac{x^c}{x^a}\right)^{(c+a)}=1$

Sol :

Using property : $\dfrac{a^m}{a^n}=a{(m-n)}$

⇒{x^(a-b)}^(a+b) × {x^(b-c)}^(b+c) × {x^(c-a)}^(c+a)

Using property : (a^m)ⁿ=a^m×n

⇒x^(a-b)×(a+b) × x^(b-c)×(b+c) × x^(c-a)×(c+a)

Using property : (a+b)(a-b)=a²-b²

⇒$x^{a^2-b^2} \times x^{b^2-c^2} \times x^{c^2-a^2}$

Using property : a^m×aⁿ=a^(m+n)

⇒$x^{a^2-b^2+b^2-c^2+c^2-a^2}$

⇒x⁰

Using property : a⁰=1

⇒1

Hence proved

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(ii) $\dfrac{x^{a+b}\times x^{b+c}\times x^{c+a}}{(x^a \times x^b \times x^c)^2}=1$

Sol :

Using property : a^m×aⁿ=a^(m+n)

⇒$\dfrac{x^{a+b+b+c+c+a}}{(x^{a+b+c})^2}$

⇒$\dfrac{x^{2a+2b+2c}}{(x^{a+b+c})^2}$

⇒$\dfrac{\not{(x^{2a+2b+2c})}}{\not{(x^{2a+2b+2c})}}$

⇒1

Hence proved

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Question 12

Find x such that

(i) $\left(\dfrac{7}{4}\right)^{-3} \times \left(\dfrac{7}{4}\right)^{-5}=\left(\dfrac{7}{4}\right)^{3x-2}$

Sol :

Using property : a^m×aⁿ=a^(m+n)

⇒$\left(\dfrac{7}{4}\right)^{-3-5}=\left(\dfrac{7}{4}\right)^{3x-2}$

⇒$\left(\dfrac{7}{4}\right)^{-8}=\left(\dfrac{7}{4}\right)^{3x-2}$

On comparing both sides

⇒-8=3x-2

Transposing -2

⇒-8+2=3x

Transposing 3

⇒$x=\dfrac{-6}{3}=-2$

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(ii) $\left(\dfrac{125}{8}\right)\times \left(\dfrac{125}{8}\right)^x=\left(\dfrac{5}{2}\right)^{18}$

Sol :

Using property : a^m×aⁿ=a^(m+n)

⇒$\left(\dfrac{125}{8}\right)^{x+1}=\left(\dfrac{5}{2}\right)^{18}$

⇒$\left(\dfrac{5^3}{2^3}\right)^{x+1}=\left(\dfrac{5}{2}\right)^{18}$

⇒$\left(\dfrac{5}{2}\right)^{3(x+1)}=\left(\dfrac{5}{2}\right)^{18}$

Comparing both sides

⇒3(x+1)=18

⇒3x+3=18

⇒3x=18-3

⇒$x=\dfrac{15}{3}=5$

 

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Question 13

Find the value of x such that

(i) $3^{2x-1}=\dfrac{1}{27^{x-3}}$

Sol :

⇒$3^{2x-1}=\dfrac{1}{3^{3(x-3)}}$

⇒3^2x-1=3^-3x+9

On comparing both sides

⇒2x-1=-3x+9

⇒3x+2x=9+1

⇒5x=10

⇒$x=\dfrac{10}{5}=2$

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(ii) $\left(\dfrac{p}{q}\right)^{3x+2}=\left(\dfrac{q}{p}\right)^{2-x}$

Sol :

⇒$\left(\dfrac{p}{q}\right)^{3x+2}=\left(\dfrac{p}{q}\right)^{-2+x}$

On comparing both sides

⇒3x+2=-2+x

⇒3x-x=-2-2

⇒2x=-4

⇒$x=\dfrac{-4}{2}=-2$

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